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Friday, 31 January 2014

Separable channels decrease the entaglement of formation

The entanglement of formation of a density operator $\rho\in\Density(\X^{A}\otimes\X^{B})$ is defined as \[ E_{f}(\rho) = \inf\Biggl\{\sum_{a\in\Sigma} p(a) E(u_a u_a^{\ast}) \,:\, \rho = \sum_{a\in\Sigma} p(a) u_a u_a^{\ast} \Biggr\}, \] where $E(u u^{\ast}) = S(\tr_{\X^{B}}(u u^{\ast}))$ denotes the entanglement entropy of the pure state $u u^{\ast}$ and the infimum is over all expressions of $\rho$ of the given form, where $\Sigma$ is any alphabet, $p\in\P(\Sigma)$ is a probability vector, and $\{u_a\,:\,a\in\Sigma\} \subset \X^{A}\otimes\X^{B}$ is a collection of unit vectors.Theorem:

For every choice of complex Euclidean spaces $\X^{A}$, $\X^{B}$, $\Y^{A}$, and $\Y^{B}$, every density operator $\rho\in\Density(\X^{A}\otimes\X^{B})$, and every separable channel $\Phi\in\SepC(\X^{A},\Y^{A}: \X^{B},\Y^{B})$, it holds that \[ E_{f}(\Phi(\rho)) \leq E_{f}(\rho). \]

Consider the channel $\Psi_{ab}\in\SepC(\X^{A},\Y^{A}: \X^{B},\Y^{B})$ defined by\[\Psi_{ab}(X)=C_bXC_b^\ast+(\tr(X)-\tr(C_bXC_b^\ast))\sigma,\]for some arbitrary $\sigma\in\Density(\Y^A\otimes\Y^B)$. Then $\Psi_{ab}$ is indeed a channel since it is completely positive because it is defined in terms of $C_b$ and $\Phi$ is assumed to be completely positive. Likewise, $\Psi_{ab}$ is separable since $\Phi$ is separable. Moreover, $\Psi_{ab}$ is trace preserving since\[\begin{align*}\tr(\Psi_{ab}(X))&=\tr(C_bXC_b^\ast)+(\tr(X)-\tr(C_bXC_b^\ast))\tr(\sigma) \\&=\tr(C_bXC_b^\ast)+\tr(X)-\tr(C_bXC_b^\ast) \\&=\tr(X).\end{align*}\]By construction $\Psi_{ab}(u_au_a^\ast)=v_{ab}v_{ab}^\ast$. Therefore, by a corollary (6.36) to Nielsen's theorem it follows that for every $a\in\Sigma$ with $\rho_a^A=\tr_{\X^B}(u_au_a^\ast)$ and $\sigma_a^A=\tr_{\X^B}(v_{ab}v_{ab}^\ast)$ and $r=min\{rank(\rho_a^A),rank(\sigma^A_a)\}$ it holds that \[\lambda_1(\rho_a^A)+\dots+\lambda_1(\rho_m^A)\leq \lambda_1(\sigma_a^A)+\dots+\lambda_1(\sigma_m^A)\]for every $m\in\{1,\dots, r\}$.

Thus, the von Neummann entropy satisfies $S(\sigma_a^A)\leq S(\rho_a^A)$, which implies that the entanglement entropy also satisfies $E(v_{ab}v_{ab}^\ast)\leq E(u_au_a)$ for all $a\in\Sigma$ and $b\in\Gamma$. Then by tracing out system $B$, and taking the weighted average that is described the original state $\rho$ and using the joint convexity of the von Neumann entropy it follows that \[\begin{align*}\sum_{a\in\Sigma} p(a)\sum_{b\in\Gamma}q_a(b) E(v_{ab} v_{ab}^{\ast})\leq \sum_{a\in\Sigma} p(a) E(u_a u_a^{\ast}).\end{align*}\]